Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $n = \dfrac{-5z^2 + 50z}{2z - 20} \times \dfrac{z^2 + 12z + 35}{z^3 - 5z^2 - 50z} $
Explanation: First factor out any common factors. $n = \dfrac{-5z(z - 10)}{2(z - 10)} \times \dfrac{z^2 + 12z + 35}{z(z^2 - 5z - 50)} $ Then factor the quadratic expressions. $n = \dfrac {-5z(z - 10)} {2(z - 10)} \times \dfrac {(z + 5)(z + 7)} {z(z + 5)(z - 10)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac {-5z(z - 10) \times (z + 5)(z + 7) } {2(z - 10) \times z(z + 5)(z - 10) } $ $n = \dfrac {-5z(z + 5)(z + 7)(z - 10)} {2z(z + 5)(z - 10)(z - 10)} $ Notice that $(z + 5)$ and $(z - 10)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {-5z\cancel{(z + 5)}(z + 7)(z - 10)} {2z\cancel{(z + 5)}(z - 10)(z - 10)} $ We are dividing by $z + 5$ , so $z + 5 \neq 0$ Therefore, $z \neq -5$ $n = \dfrac {-5z\cancel{(z + 5)}(z + 7)\cancel{(z - 10)}} {2z\cancel{(z + 5)}\cancel{(z - 10)}(z - 10)} $ We are dividing by $z - 10$ , so $z - 10 \neq 0$ Therefore, $z \neq 10$ $n = \dfrac {-5z(z + 7)} {2z(z - 10)} $ $ n = \dfrac{-5(z + 7)}{2(z - 10)}; z \neq -5; z \neq 10 $